"I don't think outside the box; I think of what I can do with the box." - Henri Matisse
Norm,
While this is a possible 'pencil-and-paper' problem, I was wondering is computer code/scripts could be posted on the forum?
If so, here is a link to the script I wrote. Please feel free to remove it, or even critique it as much as possible:
https://ideone.com/dnaKM2
I am not going to take a shot at the expected value but the game is one you should play. You pay $250 and you get at least $200 back and lose $50. Every other combination has you winning $50 plus $0, $100, 200... to $1000. I will assume a single decimal digit so you draw 0.1, 0.2, ... to 0.9 or 9 possible numbers. The between 0 and 1 does not say inclusive. It does not specify whether you draw with or without replacement but one of the examples draws the same number twice so we can assume it is with replacement or an infinite series of numbers. After 2 draws you will have lost $50 36 times and still be drawing 45 times and win at least $50 for those times.
I can't prove it, but I'd say play it.
My thinking is, if half the time there are 2 picks and half the time there are 3, then you're breaking even. Average pick is 0.5 (assuming 0.1, 0.2..., 0.9). Since the sum must EXCEED 1.0, after 2 picks your average result will be 1.0, meaning exceeding 1.0 after 2 picks cannot be 50%. Since it therefore must be less than 50%, then you're playing a winning game -- even IF the max layout was only $300.
Or
If getting 2 picks to exceed 1.0 occurred 50% of the time, and since we know it is possible to get a payout which exceeds $300, then the game must be +EV (since it'd be break even if $200 vs $300 was 50/50.....now it's like $200 @ 50%, $300 at 49%, and $400 @ 1%.....well, that's not what it is, but even if that's what it was, it'd still be +EV).
There'd be 81 ways for the first two picks to occur, assuming 0-1 exclusive (not 0 and not 1).
First is 9, there are 8 ways to exceed 1.0
First is 8, there are 7 ways to exceed 1.0
7....6
6...5
5..4
4..3
3...2
2...1
1....0
36 ways out of 81 to lose $50. 45 ways to win at least $50. That's +EV.
0 to 1 including 0 and 1:
121 ways to pick 2 numbers
First is 0....0 ways to exceed 1.0
First is 1....1 way
2...2
3....3
4....4
1.0....10 ways
1+2+3+4+5+6+7+8+9+10 = (11*5) = 55
55 ways to lose $50 and 66 ways to win at least $50.....that's also +EV.
Edit: assume those are decimals above (8 meaning 0.8).
Edit edit: if this wants to pick a random num with more precise digits like if it can choose 0.11562 or whatever......then I got no idea how to "show my work".
Also, idk if my logic above is sound or not. But this seems far too easy of an answer to be correct.
Last edited by RS; 03-28-2016 at 08:19 AM.
"Everyone wants to be rich, but nobody wants to work for it." -Ryan Howard [The Office]
RS you went in much more detail than I did, but here was my first thought no calculator thoughts:
From the context of the question there exists a random number generator pulling numbers between 0 and 1 (.1,... .9, etc). It will add the numbers together and whenever the number is OVER 1, it stops. This is the key factor in the wording of the riddle, in my opinion. The value of the result must be GREATER THAN 1 for the game to stop. For every draw you earn $100. If this is the case the mean of these numbers would be .5, and the average number of draws to meet the requirements would be 3, because the result value of 1 requires another draw to get GREATER THAN 1. Thus, on average, you would require 3 draws, making $300. There is a $250 entry fee, so your Expected Value each time you play the game would be $50.
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Originally Posted by DSchles
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